118 CHEMICAL ENGINEERING
most important. From the HQ points the water horsepower can be calculated and by comparing this with the observed brake horsepower the pump efficiency can be computed and the EQ or efficiency-capacity curve added to the diagram.
These three curves are shown in Fig. 11. The test is started with the discharge valve wide open, which produces the maximum capacity against practically no head. This condition is indicated by point Q, corresponding, in the case illustrated, to a capacity of 4,400 gal. per minute. The discharge valve is now gradually closed, whereupon the capacity diminishes and the head rises until it reaches a maximum value at H 193 ft. From this point on it decreases until, when the discharge valve is closed the shut-off head, Hs = 189 ft., is reached at the point of no capacity. Under both conditions given by the two extremes, viz.: Head = 0 and Capacity = 0, the Efficiency = 0. Between these extremes the efficiency curve rises to a maximum value, in this case, E = 74 per cent. Generally this maximum efficiency is only obtained for one set of conditions, in this case, when H = 168 ft. and Q = 2600 gal. per minute. For any other condition the efficiency will be less and for this reason an efficiency of probably 70 per cent would be guaranteed. This guaranteed efficiency would then hold good for a range of conditions, indicated by the line between points A and C. By partly closing the discharge valve the normal head in the column pipe may be maintained for quite a range in capacity. Thus, if the normal head was 145 ft. a capacity of 3,300 gal. per minute (point C) could be obtained, but by throttling the discharge it would also be possible to run at point G} giving a head of 182 ft. in the pump while maintaining the head of 145 ft. in the column without dropping below the efficiency of 70 per cent. Under these conditions the capacity would then be 1,800 gal. per minute so that by merely throttling the discharge any capacity between 1,800 and 3,300 gal. per minute against a head of 145 ft. with an efficiency of 70 per cent can be had.
The three factors: speed, capacity and head are intimately related to each other, and the relation is given by the fundamental equation. The speed is represented by wtt, the capacity is proportional to vr, and H is the head pumped against. In the following a few rules will be given for analyzing the performance of a pump: In a given impeller the vane angle aa is known, and the factors ua, vr, H and Eh can be computed from the test results by using the following formulas: ua - dairn -r- 12 X 60 = dan -s-229.2; vr « Q -r- area = Q + 12(daTr tSa * sin a,a)ba; Eh ~ Hg/(ua q)ua where da = external diameter of impeller, inches, n revolutions per minute, Q = capacity in cubic inches per second, z number of vanes, sa thickness of vanes, inches at exit, g = 32.2 ft. per second, q = the horizontal component of the velocity va in feet per second = vr -? tan c*a, and ba the width of impeller at exit, inches.
Example.The pump to be tested has an impeller of the following dimensions: da = 21.375 in., ba = 1J{6 in., z = 6, s« = 0.25 in., ota = 18.5° and is running at the speed of 1,100 r.p.m. The test results are plotted in Fig. 11. The maximum head obtained was Hm 193 ft., the maximum efficiency E = 74 per cent, arid the head at maximum efficiency was H = 168 ft. The capacity at the point of maximum efficiency was Q = 2,600 gal. per minute = 10,010 cu. in. per second. The maximum capacity was Qm = 4,400 gal. per minute. By substituting these values based on actual test in the above formulas, we can compute the various factors composing the fundamental equation, viz.:
ua = 21.375 X 1,100 -T- 229.2 = 102.6 ft. per second
vr = 10,010 -s- 12(21.375 TT - 6 X 0.25 -f- sin 18.5°) 1.56 = 8.56 feet per second. q = 8.56 -T- tan 18.5° = 25.6 feet per second Ua - q = 102.6 - 25,6 - 77 ft. per second
Eh = 32.2 X 168 -* 102.6 X 77 - 68.5 per cent.